[原创]Light 1045 Digits of Factorial 【数论】

2016-06-30 15:54:37 Tabris_ 阅读数：241

博客爬取于2020-06-14 22:44:17
以下为正文

版权声明：本文为Tabris原创文章，未经博主允许不得私自转载。
https://blog.csdn.net/qq_33184171/article/details/51791054

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=120197#problem/U

Digits of Factorial
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit

Status
Description
Factorial of an integer is defined by the following function

f(0) = 1
f(n) = f(n - 1) * n, if(n > 0)

So, factorial of 5 is 120. But in different bases, the factorial may be different. For example, factorial of 5 in base 8 is 170.

In this problem, you have to find the number of digit(s) of the factorial of an integer in a certain base.

Input
Input starts with an integer T (≤ 50000), denoting the number of test cases.

Each case begins with two integers n (0 ≤ n ≤ 106) and base (2 ≤ base ≤ 1000). Both of these integers will be given in decimal.

Output
For each case of input you have to print the case number and the digit(s) of factorial n in the given base.

Sample Input
5
5 10
8 10
22 3
1000000 2
0 100
Sample Output
Case 1: 3
Case 2: 5
Case 3: 45
Case 4: 18488885
Case 5: 1

题目大意　　：　　就是求n!在base进制下表示 不算前导0的话 有多少位

多少位的话 直接取log即可
但是n大了 long long 会爆 所以
拆成
log(n!)=log(n)+log(n-1)+ … （以上底数为base）.

因为计算机中只有底数为十的对数函数 所以 表示成lg(n!)/lg(base) 就行

最后n==0 的时候要特判一下.

上面就是题解了
但是 我做的时候出现了 迷之RE+迷之WA

RE ： printf(“Case %d: %.0llf\n”,++p,f[n]/log(base)+0.5);
这么输出居然是RE

于是又换了一种

WA：printf(“Case %d: %d\n”,++p,(int)(f[n]/log(1.0*base)+0.5));

这样居然是WA ？我就不明白了

最后还是看别人代码 用的ceil() 才过的 但这到底是为什么啊啊啊啊啊啊啊啊啊
最终过得姿势 ::: printf(“Case %d: %d\n”,++p,(int)ceil(f[n]/log(1.0*base)))

附本题代码

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# include <stdio.h>
# include <stdlib.h>
# include <string.h>
# include <limits.h>
# include <malloc.h>
# include <ctype.h>
# include <math.h>
# include <string>
# include <iostream>
# include <algorithm>
# include <map>
# include <vector>
# include <set>

using namespace std;

# define LL long long int

double f[1001110];


int main()
{
    f[0]=0;
    for(int i=1;i<=1000010;i++)
    {
        f[i]=f[i-1]+log(1.0*i);
    }

    int t,p=0,n,base;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&base);

        if(n==0)
        {
            printf("Case %d: 1\n",++p);
            continue;
        }
        printf("Case %d: %d\n",++p,(int)ceil(f[n]/log(1.0*base)));
    }
    return 0;
}